# Finding a Tangent Plane on a Surface Examples 1

Recall from the Finding a Tangent Plane on a Surface page that if $z = f(x, y)$ is a two variable real-valued function, then the equation of the tangent plane at a point $P(x_0, y_0)$ on the surface is given by the following formula:

(1)We will now look at some more examples of finding tangent planes to surfaces.

## Example 1

**Find the equation of the tangent plane that passes through point generated by $(1, 0) \in D(f)$ on the surface $f(x, y) = x^2e^y + \frac{1}{2}x^3y$.**

The partial derivatives of this function are $f_x(x, y) = 2xe^y + \frac{3}{2} x^2 y$ and $f_y(x, y) = x^2 e^y + \frac{1}{2} x^3$, so $f_x(1, 0) = 2$ and $f_y(1, 0) = \frac{3}{2}$ and the function evaluated at the point $(x_0, y_0) = (1, 0)$ is $z_0 = f(1, 0) = 1$, and so the equation of the tangent plane that passes through $(1, 0, 1)$ on this surface is:

(2)## Example 2

**Find the equation of the tangent plane that passes through the point generated by $(1, 1) \in D(f)$ on the surface $f(x, y) = \ln (x^2 + \frac{1}{2}e^x y^2)$.**

The partial derivatives of this function are $f_x(x, y) = \frac{2x + \frac{1}{2}e^x y^2}{x^2 + \frac{1}{2}e^x y^2}$ and $f_y(x, y) = \frac{e^x y}{x^2 + \frac{1}{2}e^x y^2 }$, so $f_x(1, 1) = \frac{2 + \frac{e}{2}}{1 + \frac{e}{2} }$ and $f_y(1, 1) = \frac{e}{ 1 + \frac{e}{2}}$ and the function evaluated at the point $(x_0, y_0) = (1, 1)$ is $z_0 = f(1, 1) = \ln \left (1 + \frac{e}{2} \right )$ and so the equation of the tangent plane that passes through $\left ( 1, 1, \ln \left (1 + \frac{e}{2} \right ) \right )$ on this surface is:

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